Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{5z - 35}{-z^2 + 49} \times \dfrac{-5z^2 - 10z + 175}{z + 9} $
Explanation: First factor out any common factors. $t = \dfrac{5(z - 7)}{-(z^2 - 49)} \times \dfrac{-5(z^2 + 2z - 35)}{z + 9} $ Then factor the quadratic expressions. $t = \dfrac {5(z - 7)} {-(z + 7)(z - 7)} \times \dfrac {-5(z + 7)(z - 5)} {z + 9} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {5(z - 7) \times -5(z + 7)(z - 5) } { -(z + 7)(z - 7) \times (z + 9)} $ $t = \dfrac {-25(z + 7)(z - 5)(z - 7)} {-(z + 7)(z - 7)(z + 9)} $ Notice that $(z + 7)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {-25\cancel{(z + 7)}(z - 5)(z - 7)} {-\cancel{(z + 7)}(z - 7)(z + 9)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $t = \dfrac {-25\cancel{(z + 7)}(z - 5)\cancel{(z - 7)}} {-\cancel{(z + 7)}\cancel{(z - 7)}(z + 9)} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $t = \dfrac {-25(z - 5)} {-(z + 9)} $ $ t = \dfrac{25(z - 5)}{z + 9}; z \neq -7; z \neq 7 $